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3 years ago

Refer to table 28-6. what is the u-2 measure of labor underutilization?

Mathematics

1 answer:

mafiozo [28]3 years ago

4 0

Bvgdgmhmgjm,jh,,,,,,,,,,,,,,,,,

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Stacy goes to the county fair with her friends . The total cost of ride tickets is given by the equation c=3.5t , where c is the

madreJ [45]

Answer:

Step-by-step explanation:

Step one:

given data

we are given that the linear function for the cost is c=3.5t

c is the cost and

t is the number of tickets.

We are told that t=15, to find c, let us put the value of t in the linear function for the cost

$c=3.5(15)\\\\c=52.5$

<u>This shows that 15 tickets will cost $52.5</u>

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3 0

2 years ago

2 400 000 in standard form

Archy [21]

It's going to be 2.4 * 10 to the power of 6.

Hope this helped =)

3 0

3 years ago

Wildlife resort has eight elephant cabs born during the summer and now has 31 total how many elephants were ending was there bef

Kipish [7]

The answer would be 23. 31-8=23

4 0

3 years ago

Which two equations would be represented as parallel lines?

Mariana [72]

Answer:

The first and third options are parallel.

Step-by-step explanation:

They both have a slope of 2.

Pls mark brainliest. Thanks!

4 0

3 years ago

Read 2 more answers

A four year old is going to spin around with his arms stretched out 100 times. From past experience, his father knows it takes a

Sati [7]

Answer:

$P(X \geq0.55) \leq 0.22$

Step-by-step explanation:

Using central Limit Theorem (CLT), The sum of 100 random variables;

$Y=X_1+X_2+...+X_{100}$ is approximately normally distributed with

Y ~ N (100 × $\frac{1}{3^2}$ ) = N ( 50, $\frac{100}{9}$ )

The approximate probability that it will take this child over 55 seconds to complete spinning can be determined as follows;

N ( 50, $\frac{100}{9}$ )

$P(Y>55) =P(Z>\frac{55-50}{10/3})$

$P(Y>55) =P(Z>1.5)$

$P(Y>55) =\phi (-1.5)$

$P(Y>55) =0.0668$

Using Chebyshev's inequality:

$P(|X-\mu\geq K)\leq \frac{\sigma^2}{K^2}$

Let assume that X has a symmetric distribution:

Then:

$2P(X-\mu\geq K)\leq) \frac{\sigma^2}{K^2}$

$2P(X \geq \mu+K)\leq) \frac{\sigma^2}{K^2}$

$2P(X\geq0.5+0.05)\leq \frac{\frac{1}{\frac{3^2}{100} } }{0.05^2}$ where: ( $\sigma^2 = \frac{1}{3^2/100}$ )

$P(X \geq0.55) \leq 0.22$

6 0

3 years ago